Attenuation in fibers, absorption and scattering losses, bending losses

1. COURSE: OPTICAL COMMUNICATION (EC317)
UNIT-II
M. RAJARAO

2. DISTORTION IN OPTICAL FIBERS
▪ When an optical signal is transmitted on an optical fiber, the signal is
distorted due to two phenomena: dispersion and attenuation.
Distortion in optical fiber
Dispersion
▪ Intramodal dispersion
▪ Material dispersion
▪ Waveguide dispersion
▪ Inter modal dispersion
Attenuation
Material attenuation
▪ Absorption
▪ Scattering
Bending loss Radiation

3. DISPERSION
▪ The light is launched into an optical fiber in the form of a light pulse to transmit the
information
▪ The broadening of the transmitted light pulses as they travel along the fiber is known
as dispersion.
▪ the overlap between adjacent pulses, creating
errors in the receiver output (i.e., ISI),
resulting in the limitation of information-
carrying capacity of a fiber.

5. CHROMATIC OR INTRAMODAL DISPERSION
▪ This type of dispersion results from the finite spectral linewidth of the optical
source. Since optical sources do not emit just a single frequency but a band of
frequencies
▪ There may be propagation delay differences between the different spectral
components of the transmitted signal. This causes broadening of each
transmitted mode and hence intramodal dispersion (group velocity dispersion).
▪ The delay differences may be caused by the dispersive properties of the
waveguide material (material dispersion) and also guidance effects within
the fiber structure (waveguide dispersion).

6. MATERIAL DISPERSION
▪ This arises due to the variations of the refractive index of the core material as a
function of wavelength.
▪ This refractive index property causes a wavelength dependence of the group
velocity of a given mode; that is, pulse spreading occurs when different
wavelengths follow the same path.
▪ The group velocity is given by 𝑣𝑔 =
𝑑𝜔
𝑑𝛽
; where the propagation constant 𝛽 =
2𝜋𝑛1 𝜆
𝜆
and ω =
2𝜋𝑐
𝜆
▪ A time delay or group delay in the direction of the propagation 𝑡𝑔 =
𝐿
𝑣𝑔
𝑡𝑔 =
𝐿
𝑣𝑔
= 𝐿
𝑑𝛽
𝑑𝜔
𝑜𝑟
𝐿
𝑐
𝑑𝛽
𝑑𝑘

7. MATERIAL DISPERSION
𝑑𝛽
𝑑𝜆
= −
2𝜋
𝜆2
𝑛1 − 𝜆
𝑑𝑛1
𝑑𝜆
𝑎𝑛𝑑
𝑑𝜔
𝑑𝜆
= −
2𝜋𝑐
𝜆2
▪ Therefore
𝑡𝑔 = 𝐿
𝑑𝛽/𝑑𝜆
𝑑𝜔/𝑑𝜆
=
𝐿
𝑐
𝑛1 − 𝜆
𝑑𝑛1
𝑑𝜆
▪ If the spectral width of the optical source is not too wide, then the delay difference per unit
wavelength along the propagation path is approximately
𝑑𝑡𝑔
𝑑𝜆
.
▪ For spectral components which are 𝛿𝜆 apart, symmetrical around center wavelength, the total
delay difference 𝛿𝜏 over a distance L is:
𝛿𝜏 =
𝑑𝑡𝑔
𝑑𝜆
𝛿𝜆 =
𝐿
𝑐
𝜆
𝑑2𝑛1
𝑑𝜆2
𝛿𝜆
▪ The material dispersion of optical fibers is quoted in terms of the material dispersion parameter
Dmat given by
𝐷𝑚𝑎𝑡 =
1
𝐿
𝛿𝜏
𝛿𝜆
=
𝜆
𝑐
𝑑2𝑛1
𝑑𝜆2
Dmat has the units of ps nm–1 km–1.

8. ▪ In the case of optical pulse, if the spectral width of the optical source is
characterized by its rms value of the Gaussian pulse 𝜎𝜆, the pulse
spreading over the length of L, 𝜎𝑚 can be well approximated by:
𝜎𝑚𝑎𝑡 =
𝑑𝑡𝑔
𝑑𝜆
𝜎𝜆 = 𝐿𝐷𝑚𝑎𝑡𝜎𝜆

10. WAVEGUIDE DISPERSION
▪ The degree of waveguide dispersion depends on the fiber design.
▪ Waveguide dispersion usually can be ignored in multimode fibers, but its effect
is significant in single-mode fibers.
▪ This dispersion arises because the difference in core-cladding spatial power
distributions, together with the speed variations of the various wavelengths,
causes a change in propagation velocity for each spectral component.
▪ Waveguide dispersion is due to the dependency of the group velocity of the
fundamental mode as well as other modes on the V number.
▪ In order to calculate waveguide dispersion, we consider that n is not dependent
on wavelength. Defining the normalized propagation constant b as:
𝑏 =
𝛽
𝑘
2
− 𝑛2
2
𝑛1
2 − 𝑛2
2

11. WAVEGUIDE DISPERSION
For small values of the index difference Δ = (n1 - n2)/n1, can be approximated by
𝑏 ≈
𝛽
𝑘
− 𝑛2
𝑛1 − 𝑛2
⇒ 𝛽 ≈ 𝑘𝑛2 𝑏∆ + 1
the group delay twg arising from waveguide dispersion is 𝑡𝑔 =
𝐿
𝑣𝑔
=
𝐿
𝑐
𝑑𝛽
𝑑𝑘
𝑡𝑔 =
𝐿
𝑐
𝑛2 1 + ∆
𝑑 𝑘𝑏
𝑑𝑘
=
𝐿
𝑐
𝑛2 1 + ∆ 𝑏 + 𝑘
𝑑𝑏
𝑑𝑘
=
𝐿
𝑐
𝑛2 1 + ∆ 𝑏 + 𝑘
𝑑𝑏
𝑑𝑉
𝑑𝑉
𝑑𝑘
𝑉 = 𝑘𝑎𝑛1 2∆ ⇒
𝑑𝑉
𝑑𝑘
= 𝑎𝑛1 2∆=
𝑉
𝑘
𝑡𝑔 =
𝐿
𝑐
𝑛2 1 + ∆
𝑑 𝑏𝑉
𝑑𝑉
The waveguide dispersion is
𝐷𝑤𝑔 =
1
𝐿
𝑑𝑡𝑔
𝑑𝜆
=
𝑛2∆
𝑐
𝑑
𝑑𝜆
𝑑 𝑏𝑉
𝑑𝑉
𝑑𝑉
𝑑𝑉
=
𝑛2∆
𝑐
𝑉
𝜆
𝑑2 𝑏𝑉
𝑑𝑉2

13. ▪ The wavelength at which the first-order dispersion
is zero λ0 may be extended to wavelengths of 1.55
μm and beyond by a combination of three
techniques. These are:
▪ lowering the normalized frequency (V) for the
fiber;
▪ increasing the relative refractive index
difference Δ for the fiber;
▪ suitable doping of the silica with germanium.

14. INTER-MODAL DISPERSION
▪ The light rays travel different paths inside the core of an optical fiber because
different light rays are incident on the tip of the optical fiber at different angles
within the acceptance cone and lead to the broadening of the actual time-width of
the pulse. This phenomenon is called as inter-modal dispersion
▪ Intermodal dispersion or modal delay appears only in multimode fibers.

15. INTER-MODAL DISPERSION
▪ This broadening is simply obtained from ray tracing and for a fiber of length L
is given by
Δ𝑇 = 𝑇𝑚𝑎𝑥 − 𝑇𝑚𝑖𝑛
𝑇𝑚𝑖𝑛 =
𝐿
𝑐
𝑛1 𝑎𝑛𝑑 𝑇𝑚𝑎𝑥 =
𝐿
𝑐 cos 𝜃
𝑛1
sin 𝜙𝑐 = cos 𝜃 =
𝑛2
𝑛1
Δ𝑇 =
𝐿
𝑐
𝑛1
𝑛1
𝑛2
− 1 ≈
𝐿
𝑐
𝑛1Δ
The rms pulse broadening due to
intermodal dispersion

16. PROBLEMS
A 6 km optical link consists of multimode step index fiber with a core refractive index of
1.5 and a relative refractive index difference of 1%. Estimate:
(a) the delay difference between the slowest and fastest modes at the fiber output;
(b) the rms pulse broadening due to intermodal dispersion on the link;
(c) the maximum bit rate that may be obtained without substantial errors on the link
assuming only intermodal dispersion
Sol:
(a) The delay difference
∆𝑇 =
𝐿
𝑐
𝑛1Δ = 6 × 1.5 ×
0.01
3 × 105
= 300 𝑛𝑠
(b) The rms pulse broadening due to intermodal dispersion
𝜎𝑠 =
𝐿
2 3𝑐
𝑛1Δ = 86.7 𝑛𝑠
(c) The maximum bit rate may be estimated in two ways:
𝐵𝑇,𝑚𝑎𝑥 =
1
2∆𝑇
= 1.7 𝑀𝑏𝑝𝑠 or 𝐵𝑇,𝑚𝑎𝑥 =
0.2
𝜎𝑠
= 2.3 𝑀𝑏𝑝𝑠

17. PROBLEMS
▪ A glass fiber exhibits material dispersion given by |λ2(d2n1/dλ2) | of 0.025.
Determine the material dispersion parameter at a wavelength of 0.85 μm, and
estimate the rms pulse broadening per km for a good LED source with an rms
spectral width of 20 nm at this wavelength.
Sol:
Given that 𝜆2 𝑑2𝑛1
𝑑𝜆2 = 0.025
Operating wavelength λ = 850 nm
RMS spectral width 𝜎𝜆 = 20 𝑛𝑚
The material dispersion
𝐷𝑚𝑎𝑡 =
𝜆
𝑐
𝑑2
𝑛1
𝑑𝜆2
=
1
𝑐𝜆
𝜆2
𝑑2
𝑛1
𝑑𝜆2
=
0.025
3 × 105 × 850
= 98.04 ps nm−1 km−1
The rms pulse broadening per km
𝜎𝑚𝑎𝑡
𝐿
= 𝐷𝑚𝑎𝑡𝜎𝜆 = 1.96 𝑛𝑠/𝑘𝑚

18. PROBLEMS
▪ A manufacturer’s data sheet lists the material dispersion Dmat of a GeO2-
doped fiber to be 110 ps/(nm-km) at a wavelength of 860 nm. Find the rms
pulse broadening per km due to material dispersion if the optical source is a
GaAlAs LED that has a spectral width of 40 nm at an output wavelength of
860 nm.

19. ATTENUATION
▪ Signal attenuation (fiber loss) largely determines the maximum repeaterless
separation between optical transmitter & receiver (i.e., maximum
transmission distance between a transmitter and a receiver)
▪ Signal attenuation is usually expressed in the logarithmic unit of the decibel.
▪ The decibel, which is used for comparing two power levels, may be defined
as the ratio of the input (transmitted) optical power Pi into a fiber to the
output (received) optical power Po from the fiber
𝑃𝑜𝑤𝑒𝑟 𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑑𝐵𝑠 = −10 log10
𝑃𝑜
𝑃𝑖
= 10 log10
𝑃𝑖
𝑃𝑜
▪ In optical fiber communications the attenuation is usually expressed in
decibels per unit length (i.e. dB/km)
𝛼 =
1
𝐿
10 log10
𝑃𝑖
𝑃𝑜
𝑖𝑛 𝑑𝐵/𝑘𝑚

20. EXAMPLE
When the mean optical power launched into an 8 km length of fiber is 120 μW, the mean optical
power at the fiber output is 3 μW. Determine:
(a) the overall signal attenuation or loss in decibels through the fiber assuming there are no
connectors or splices;
(b) the signal attenuation per km for the fiber.
(c) the overall signal attenuation for a 10 km optical link using the same fiber with splices at 1 km
intervals, each giving an attenuation of 1 dB;
Sol:
(a) The overall signal loss/attenuation
𝛼 𝑑𝐵 = 10 log10
𝑃𝑖
𝑃
𝑜
= 16 𝑑𝐵
(b) The signal attenuation per km
𝛼 𝑑𝐵/𝑘𝑚 =
16
8
= 2 𝑑𝐵/𝑘𝑚
(c) the link also has nine splices (at 1 km intervals) each with an attenuation of 1 dB. Therefore, the
loss due to the splices is 9 dB. Hence, the overall signal attenuation for the link is: 29 dB

21. ATTENUATION IN FIBERS
▪ A number of mechanisms are responsible for the signal attenuation within
optical fibers. These mechanisms are influenced by the material composition,
the preparation and purification technique, and the waveguide structure.
▪ The basic attenuation mechanisms in a fiber are material absorption,
material scattering (linear and nonlinear scattering), curve and micro-
bending losses, mode coupling radiation losses and losses due to leaky modes.
There are also losses at connectors and splices

22. MATERIAL ABSORPTION LOSS
▪ Material absorption is related to the material composition and the fabrication process
for the fiber, which results in the dissipation of some of the transmitted optical power as
heat in the waveguide.
▪ Absorption by atomic defects in the glass composition (i.e., imperfections in the atomic
structure of the fiber material).
▪ these defects include missing molecules, high-density clusters of atom groups, or oxygen
defects in the glass structure
▪ Absorption is caused by the following mechanisms:
▪ Intrinsic absorption
▪ Extrinsic absorption

23. INTRINSIC AND EXTRINSIC ABSORPTION LOSS
▪ Intrinsic absorption caused by the interaction with one or more of the major components of the
glass or fiber material.
▪ electronic absorption bands (associated with the band gaps of the amorphous glass materials)
in the ultra violet (UV) region and from atomic vibration bands in the near-infrared (IR)
region
▪ The UV loss contribution in dB/km at any wave length (given in µm) can be expressed
empirically (derived from observation or experiment) as a function of the mole fraction x
𝛼𝑢𝑣 =
154.2𝑥
46.6𝑥 + 60
× 10−2 exp
4.63
𝜆
▪ In the near-infrared region above 1.2 µm,
𝛼𝐼𝑅 = 7.81 × 1011 exp
−48.48
𝜆
▪ Extrinsic absorption by impurity atoms in the glass material.

25. SCATTERING LOSS
▪ Small (compared to wavelength) variation in material density, chemical composition, and
structural inhomogeneity scatter light in other directions and absorb energy from guided
optical wave.
▪ The essential mechanism is the Rayleigh scattering. Since the black body radiation classically
is proportional to 𝜆−4, the attenuation coefficient due to Rayleigh scattering is approximately
proportional to 𝜆−4
.
e
Temperatur
:
,
JK
10
3806
.
1
Js,
10
626
.
6 -1
23
34
T
k
h B
−
−

=

=
where p is the photo-elastic coefficient
βT is the isothermal compressibility of the material

27. BENDING LOSS
▪ An optical fiber tends to radiate propagating power
whenever it is bent. Two types of bends
▪ macro-bends with radii much larger than the fiber
diameter and
▪ random micro-bends of the fiber axis that may arise
because of faulty cabling.
▪ Every guided core mode has a modal electric-field
distribution that has a tail extending into the cladding.
This evanescent field decays exponentially with
distance from the core.
▪ the optical energy in the tail beyond rc is lost through
radiation
▪ The amount of optical radiation from a bent fiber depends on the field strength at rc and on the
radius of curvature R.
▪ Since higher-order modes are bound less tightly to the fiber core than lower order modes, the
higher-order modes will radiate out of the fiber first.

28. EXAMPLE PROBLEM
▪ Consider a graded-index multimode fiber for which the index profile α = 2.0, the core index
n1= 1.480, the core-cladding index difference Δ = 0.01, and the core radius a = 25 µm. If the
radius of curvature of the fiber is R = 1.0 cm, what percentage of the modes remain in the
fiber at a 1300-nm wavelength?
Thus 42 percent of the modes remain in this fiber at a 1.0-cm bend radius.

29. MICRO-BENDING LOSS
▪ The micro-bends are caused by manufacturing defects which are in the form of non-uniformities in
the core radius or in the lateral pressure created by the cabling of the fiber. The effect of mode
coupling in multimode fibers on pulse broadening can be significant for long fibers.
▪ Micro-bend losses in single-mode fibers can also be excessive if proper care is not taken to minimize
them. One way to reduce such losses in single-mode fibers is to choose the V -value near the cut-off
value of Vc (e.g., 2.405 for the step-index profile), so that the mode energy is confined mainly to the
core.

30. EXAMPLE
▪ A certain optical fiber has an attenuation of 0.6 dB/km at 1310 nm and 0.3 dB/km at
1550 nm. Suppose the following two optical signals are launched simultaneously into
the fiber: an optical power of 150 µW at 1310 nm and an optical power of 100 µW at
1550 nm. What are the power levels in mW of these two signals at (a) 8 km and (b) 20
km?
Sol:
Given data: At λ1 = 1310 nm:
𝛼1 = 0.6 𝑑𝐵/𝑘𝑚
𝑃1 = 150 𝜇𝑊
𝑃1𝑖𝑛 𝑑𝐵𝑚 = 10 log10 150 × 10−3
= −8.24 𝑑𝐵𝑚
At λ2 = 1550 nm:
𝛼2 = 0.3 𝑑𝐵/𝑘𝑚
𝑃2 = 100 𝜇𝑊
𝑃1𝑖𝑛 𝑑𝐵𝑚 = 10 log10 100 × 10−3
= −10 𝑑𝐵𝑚
(a) At 8 km:
𝑃1 8𝑘𝑚 = 𝑃1 − 𝛼1𝐿 = −8.24 − 4.8 = −13.04 𝑑𝐵𝑚 = 49.66 𝜇𝑊
𝑃2 8𝑘𝑚 = 𝑃2 − 𝛼2𝐿 = −10 − 2.4 = −12.4 𝑑𝐵𝑚 = 57.54 𝜇𝑊
(b) At 8 km:
𝑃1 20 𝑘𝑚 = 𝑃1 − 𝛼1𝐿 = −8.24 − 12 = −20.24 𝑑𝐵𝑚 = 9.46 𝜇𝑊
𝑃2 20𝑘𝑚 = 𝑃2 − 𝛼2𝐿 = −10 − 6 = −16 𝑑𝐵𝑚 = 25.12 𝜇𝑊

31. EXAMPLE
A continuous 40-km-long optical fiber link has a loss of 0.4 dB/km.
(a) What is the minimum optical power level that must be launched into the fiber to
maintain an optical power level of 2.0 µW at the receiving end?
(b) What is the required input power if the fiber has a loss of 0.6 dB/km?
Sol:
Given that L = 40 km; α = 0.4 dB/km; Pout = 2 µW
(a) 𝑃𝑖𝑛 = 𝑃𝑜𝑢𝑡 × 10𝛼𝐿/10
= 2 𝜇𝑊 × 100.4×40
= 79.6 𝜇𝑊
(b) ) 𝑃𝑖𝑛 = 𝑃𝑜𝑢𝑡 × 10𝛼𝐿/10 = 2 𝜇𝑊 × 100.6×40 = 1.004 𝑚𝑊

32. TEXT BOOKS
▪ Gerd Keiser, Optical Fiber Communications, TMH India, Fourth Edition,
2010.
▪ Senior John M., Optical Fiber Communications, Pearson Education India,
Third Edition,2009.
▪ R.P. Khare, Fiber optics and optoelectronics, Oxford University Press 2004